Question

The complete combustion of 0.521 g of a snack bar in a calorimeter (Ccal = 6.15 kJ/°C) raises the temperature of the calorimeter by 1.29 °C.  Calculate the food value (in Cal/g) for the snack bar.

Answer 1

Answer: 3,640 cal/g

Explanation:

1) Data:

m = 0.521 g

C cal = 6.15 kJ/ °C

ΔT = 1.29°C

2) Principles and formulae:

• The calorimeter is used to measure the heat exchanged between to objects.
• As the calorimeter is isolated, the heat released by the combustion of the snack bar equals the heat absorbed by the calorimeter.
• The food value of the snack bar (in Cal/g) equals the heat realeased during its combustion, so it equals the heat absorbed by the calorimeter.
• The heat absorbed by the calorimeter = C cal × ΔT

3) Solution

• Heat released by the calorimeter = 6.15 kJ/°C × 1.29°C = 7.93 kJ

• Food value =  7.93 kJ = 7,930 J

• Conversion of units: 1.00 cal = 4.184 J

        ⇒ 7,930 J = 7,930 × 1.00 cal / 4.184 J  = 1,895cal

• Cal/g = cal / m = 1,895 cal / 0.521 g = 3,637 cal/g = 3,640 cal/g

Answer 2

The food raises the temperature of the calorimeter by 1.29 °C. The amount of calorie produced by the combustion would be: 1.29°C *  6.15 kJ/°C= 7.9335 kJ.

If the food weight is 0.521g, then the food value would be: 
7.9335 kJ/ 0.521g / (4.184kcal/kJ) =
(15.227 kJ/g) / (4.184kcal/kJ)= 3.64 kilo calorie/gram